# Duality Principle (Boolean Algebras)

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*This proof is about the Duality Principle as applied to Boolean Algebras. For other uses, see Duality Principle.*

## Theorem

Let $\struct {S, \vee, \wedge}$ be a Boolean algebra.

Then any theorem in $\struct {S, \vee, \wedge}$ remains valid if both $\vee$ and $\wedge$ are interchanged, and also $\bot$ and $\top$ are interchanged throughout the whole theorem.

## Proof

Let us take the axioms of a Boolean algebra $\struct {S, \wedge, \vee}$:

\((BA \ 0)\) | $:$ | $S$ is closed under both $\vee$ and $\wedge$ | ||||||

\((BA \ 1)\) | $:$ | Both $\vee$ and $\wedge$ are commutative | ||||||

\((BA \ 2)\) | $:$ | Both $\vee$ and $\wedge$ distribute over the other | ||||||

\((BA \ 3)\) | $:$ | Both $\vee$ and $\wedge$ have identities $\bot$ and $\top$ respectively | ||||||

\((BA \ 4)\) | $:$ | $\forall a \in S: \exists \neg a \in S: a \vee \neg a = \top, a \wedge \neg a = \bot$ |

It can be seen by inspection, that exchanging $\wedge$ and $\vee$, and $\bot$ and $\top$ throughout does not change the axioms.

Thus, what you get is a Boolean algebra again.

Hence the result.

$\blacksquare$

## Sources

- 1964: W.E. Deskins:
*Abstract Algebra*... (previous) ... (next): $\S 1.5$: Theorem $1.14$ - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next): Entry:**Boolean algebra**